# Dynamical system equilibrium point increment

I am reading through the dynamical systems theory and there is an example of a Mass-Spring system. The state equations are given by

$\displaystyle \frac{d x_1}{dx}(t) = x_2(t)$

$\displaystyle \frac{d x_2}{dx}(t) = \frac{-k}{m}x_1(t)+\frac{f(t)}{m}$

Then we find the equilibrium point by setting

$\displaystyle 0 = x_2(t)$

$\displaystyle 0 = \frac{-k}{m}x_1(t)+\frac{f(t)}{m}$

Which gives as result

$x_{eq} = \left[ \begin{array}{c} x_{1} \\ x_{2} \\ \end{array} \right] = \left[ \begin{array}{c} f/k \\ 0 \\ \end{array} \right]$

Finally, the book defines the increment with respect to the equilibrium point $\Delta x(t) = x(t) - x_{eq}$. Substracting equation (1,3) and (2,4) the result is

$\displaystyle \frac{d \Delta x_1}{dx}(t) = \Delta x_2(t)$

$\displaystyle \frac{d \Delta x_2}{dx}(t) = \frac{-k}{m}\Delta x_1(t)$

So far so good, but for the next step they get "the general solution, parametrized by the initial state" as

$\displaystyle \Delta x_1(t) = \Delta x_1(0) cos(\omega t) + \frac{\Delta x_2(0)}{\omega} sin(\omega t)$

$\displaystyle \Delta x_2(t) = -\Delta x_1(0) \omega sin(\omega t) + \Delta x_2(0)cos(\omega t)$

This result I don't understand where does it comes from, could someone give me a hint?

## 1 Answer

This is ODE business, the solution of : \begin{eqnarray*} \frac{dx_1}{dt} &=& x_2\\ \frac{dx_2}{dt} &=& -\frac{k}{m} x_1 \end{eqnarray*} is of the form you wrote. Let $\omega^2 = \frac{k}{m}$ for simplification.

A way to prove it is to differentiate the first equation and substitute the second in: $$\frac{d^2x_1}{dt^2} = \frac{dx_2}{dt} = -\omega^2 x_1$$

So the solution for $x_1$ is of the form : $$x_1(t) = A_1\cos(\sqrt{\omega}t) + A_2\sin(\sqrt{\omega}t)$$

Differentiate it to find the solution to $x_2(t)$.

Then substituting the initial conditions you find $A_1 = x_1(0)$ and $A_2 = \frac{x_2(0)}{\omega}$.